平面 p 和 q 的方程分别为： $r = (2 + \lambda - 2\mu)\,\mathbf{i} + (-3\lambda + \mu)\,\mathbf{j} + (3 - \lambda + 2\mu)\,\mathbf{k}$ 与 $r \cdot \begin{pmatrix} a \\ -1 \\ b \end{pmatrix} = 1$， 其中 a 和 b 是常数，$\lambda$ 和 $\mu$ 是参数。 直线 l 通过点 (5,4,0) 且平行于向量 $-2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$。平面 p 和 q 相交于直线 l。

1. Substitute the position vector of the known point on $l$ into the equation of plane $q$ to find $a$.

Since the line $l$ lies on the plane $q$, any point on $l$ must satisfy the equation of $q$. We are given that $l$ passes through $(5, 4, 0)$.
$\begin{pmatrix} 5 \\ 4 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} a \\ -1 \\ b \end{pmatrix} = 1$
$5(a) + 4(-1) + 0(b) = 1$
$5a - 4 = 1$
$5a = 5 \implies a = 1$

2. Use the condition that the direction vector of $l$ is perpendicular to the normal vector of $q$ to find $b$.

Since the line $l$ is entirely on the plane $q$, its direction vector must be perpendicular to the normal vector of $q$. Thus, their scalar product is zero.
$\begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ b \end{pmatrix} = 0$
$-2(1) + (-1)(-1) + 2(b) = 0$
$-2 + 1 + 2b = 0$
$2b - 1 = 0 \implies 2b = 1 \implies b = \frac{1}{2}$

1. Find the normal vector to plane $p$.

From the parametric equation of plane $p$, the two direction vectors are $\mathbf{d}_1 = \begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}$.
The normal vector to plane $p$ is their cross product:
$\mathbf{n}_p = \begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix} \times \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} (-3)(2) - (-1)(1) \\ (-1)(-2) - (1)(2) \\ (1)(1) - (-3)(-2) \end{pmatrix} = \begin{pmatrix} -5 \\ 0 \\ -5 \end{pmatrix}$
For simpler calculations, we can take a parallel vector $\mathbf{n}_p = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$.

2. Identify the normal vector to plane $q$.

Using $a = 1$ and $b = \frac{1}{2}$ from part (a), the equation of plane $q$ is $\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ \frac{1}{2} \end{pmatrix} = 1$.
The normal vector to plane $q$ is:
$\mathbf{n}_q = \begin{pmatrix} 1 \\ -1 \\ \frac{1}{2} \end{pmatrix}$
To avoid fractions, we can take a parallel vector $\mathbf{n}_q = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$.

3. Calculate the exact acute angle between the planes.

Let $\theta$ be the acute angle between the planes, which corresponds to the acute angle between their normal vectors.
$\cos \theta = \frac{|\mathbf{n}_p \cdot \mathbf{n}_q|}{|\mathbf{n}_p| |\mathbf{n}_q|} = \frac{\left| \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} \right|}{\sqrt{1^2+0^2+1^2} \sqrt{2^2+(-2)^2+1^2}}$
$\cos \theta = \frac{|2 + 0 + 1|}{\sqrt{2} \sqrt{9}} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$
Since we need the exact acute angle,
$\theta = \frac{\pi}{4} \text{ (or } 45^\circ)$

1. Find the perpendicular distance from point A to plane q.

From part (a), $a=1$ and $b=\frac{1}{2}$. The equation of plane q is $\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ 1/2 \end{pmatrix} = 1$, which can be scaled to $\mathbf{r} \cdot \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = 2$.

The perpendicular distance from $A(2, 0, 3)$ to plane q is:
$AM = \frac{|2(2) + (-2)(0) + 1(3) - 2|}{\sqrt{2^2 + (-2)^2 + 1^2}}$
$AM = \frac{|4 + 0 + 3 - 2|}{\sqrt{9}} = \frac{5}{3}$

2. Concisely evaluate the acute angle between planes p and q (result from part b).

The normal to plane p is $\mathbf{n}_p = \begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix} \times \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -5 \\ 0 \\ -5 \end{pmatrix}$.
The normal to plane q is $\mathbf{n}_q = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$.
If $\theta$ is the acute angle between the planes:
$\cos \theta = \frac{|\mathbf{n}_p \cdot \mathbf{n}_q|}{|\mathbf{n}_p| |\mathbf{n}_q|} = \frac{|-10 + 0 - 5|}{\sqrt{50} \sqrt{9}} = \frac{15}{15\sqrt{2}} = \frac{1}{\sqrt{2}}$
$\theta = 45^\circ$

3. Use geometric reasoning to deduce the shortest distance from A to l.

By substituting $\lambda = 0$ and $\mu = 0$ into the equation of plane p, we obtain the position vector $\begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}$, confirming that A lies exactly on plane p.
Line $l$ is the intersection of planes p and q, so it lies on both planes.
Let $M$ be the foot of the perpendicular from A to q, and $N$ be the foot of the perpendicular from A to $l$.
Since $l$ lies on q and $AM \perp q$, we have $AM \perp l$. By definition, $AN \perp l$.
Thus, $l$ is perpendicular to the plane $AMN$. This means $\angle ANM$ corresponds exactly to the acute angle between planes p and q, so $\angle ANM = 45^\circ$.
In the right-angled $\triangle AMN$:
$\sin 45^\circ = \frac{AM}{AN} \implies AN = \frac{AM}{\sin 45^\circ} = \frac{5/3}{1/\sqrt{2}} = \frac{5\sqrt{2}}{3}$
The shortest distance from A to $l$ is $\frac{5\sqrt{2}}{3}$.

1. Deduce the geometric relationship and angle between plane $q$ and its reflection $q'$.

From part (b), the acute angle between the mirror plane $p$ and plane $q$ is $45^\circ$.
Since plane $q'$ is the reflection of $q$ about $p$, the angle between $q$ and $q'$ is $2 \times 45^\circ = 90^\circ$.
Thus, plane $q'$ is perpendicular to plane $q$.

2. Find the normal vector of plane $q'$ using the cross product.

Since $q' \perp q$, its normal vector $\mathbf{n}_{q'}$ is perpendicular to the normal vector of $q$, $\mathbf{n}_q = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$.
Plane $q'$ also contains the intersection line $l$, so $\mathbf{n}_{q'}$ must be perpendicular to the direction vector of $l$, $\mathbf{d}_l = \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix}$.
We find $\mathbf{n}_{q'}$ using the cross product:
$\mathbf{n}_{q'} = \mathbf{n}_q \times \mathbf{d}_l = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} \times \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} -4 - (-1) \\ -2 - 4 \\ -2 - 4 \end{pmatrix} = \begin{pmatrix} -3 \\ -6 \\ -6 \end{pmatrix}$
We can scale this to get a simpler parallel normal vector: $\mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.

3. Form the cartesian equation of plane $q'$ using a point on the plane.

The line $l$ passes through the point $(5, 4, 0)$, which must therefore lie on plane $q'$.
Using the scalar product equation of a plane $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$:
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = 5(1) + 4(2) + 0 = 13$
The cartesian equation of the plane $q'$ is:
$x + 2y + 2z = 13$